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4.3Priority Sequencing Rules Priority Rules provide guidelines for the sequence in which jobs should be worked. In using this rules, job processing times and due dates are important pieces of information. Priority Rules try to minimize completion time, number of jobs in the system, and job lateness, while maximizing facility utilization.

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4.3Priority Sequencing Rules 4 types First come, first served (FCFS) Shortest processing time (SPT) Earlier due date (EDD) Critical Ratio (CR)

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4.3Priority Sequencing Rules First come, first served (FCFS) Jobs are processed in the order in which they arrive at a machine or work center. Shortest processing time (SPT) Jobs are processed according to processing time at a machine or work center, shortest job first. Earlier due date (EDD) Jobs are processed according t due date, earlier due date first. Critical Ratio (CR) Jobs are processed according to smallest ratio of time remaining until due date to processing time remaining.

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4.3.1Sequencing Rules: First Come, First Served (FCFS) Example 1 JobDays to FinishDate Promised A25 B88 C612 D410 E14 Sequence Work Time Flow Time Due Date Lateness A 2 2 5 0 B 810 8 2 C 61612 4 D 42010 E 121 417 TOTALS2169 33 SOLUTION : Five jobs are to be done at custom furniture shop: Measure of effectiveness: Flow time: is the amount of time a job spent in shop/factory Total work time/makespan: is the time needed to process given set of jobs Lateness: different between completion time and due date (if (– ve) put it zero) Note! Do all the jobs get done on time? No, Jobs B, C, D and E are going to be late

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#Average completion time = Total flow time ÷ No. of jobs #Average number of jobs = Total flow time ÷ Total job work time In the system #Average job lateness = Total late days ÷ No. of jobs #Utilization = Total job work time÷ Total flow time = in % Performance measuring formula: 4.3.1Sequencing Rules: First Come, First Served (FCFS) Example 1

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Five jobs are to be assemble in AHP Plastic Sdn. Bhd.: 4.3.1Sequencing Rules: First Come, First Served (FCFS) Example 2

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SOLUTION: 4.3.1Sequencing Rules: First Come, First Served (FCFS) Example 2

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Shortest Processing Time. Jobs with the shortest processing time are scheduled first. Jobs are sequenced in increasing order of their processing time. Shortest processing time is optimal for minimizing: Average and total flow time Average waiting time Average and total lateness 4.3.2Sequencing Rules: Shortest Processing Time (SPT)

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The steps for using this rule are : 1.Firstly, the user will input the number of jobs, the job names, the processing time and the due date of each job or use the data values given at the starting point. 2.The second step is sorting out the shortest processing time among the jobs. 3.Thirdly, calculate the flow time of each job by using the processing time. The flow time is the accumulation of processing time each job by each job. 4.3.2Sequencing Rules: Shortest Processing Time (SPT)

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Suppose we have the four jobs to the right arrive for processing on one machine 4.3.2Sequencing Rules: Shortest Processing Time (SPT) Example 1

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Average completion time28/4 = 7 days Average number of jobs in the system28/15 = 1.867 jobs Average lateness8/4 = 2days Utilization15/28 = 53.57% 4.3.2Sequencing Rules: Shortest Processing Time (SPT) Example 1 Sequence Work Time Flow Time Due Date Lateness D 1 1 4 0 C 34 60 A 4 85 3 B 7 15105 TOTAL15 28 8 Answer: Shortest Operating Time Schedule Jobs A and B are going to be late

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Job sequenceProcessing TimeDue Date A(turning) B(drilling) C(grinding) D(milling) E( facing) 6283962839 8 6 18 15 23 4.3.2Sequencing Rules: Shortest Processing Time (SPT) Example 2 A Brake Pad have 5 process that will undergo before it will be produce at a particular point in time. The jobs are labeled A, B, C, D, and E in the order that they entered the shop. The respective processing times and due dates are given in the table below. Determine the schedule by using the SPT rule.

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Solution: 4.3.2Sequencing Rules: Shortest Processing Time (SPT) Example 2

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Average completion time65/5 = 13 days Average number of jobs in the system65/28 = 2.3214 jobs Average lateness9/5 = 1.8 days Utilization28/65 = 43.08% 4.3.2Sequencing Rules: Shortest Processing Time (SPT) Example 2 #Average completion time = Total flow time ÷ No. of jobs #Average number of jobs = Total flow time ÷ Total job work time In the system #Average job lateness = Total late days ÷ No. of jobs #Utilization = Total job work time÷ Total flow time = in %

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Jobs are sequenced in increasing order of their due dates; The job with earliest due date is first, the one with the next earliest due date is second, and so on; A priority sequencing rule that specifies that the job with the earliest due date is the next job to be processed 4.3.3Sequencing Rules: Earliest Due Date (EDD)

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The steps for using this rule are : 1.Firstly, the user will input the number of jobs, the job names, the processing time and the due date of each job or use the data values given at the starting point. 2.The second step is sorting out the earliest due date among the jobs. 3.Thirdly, calculate the flow time of each job by using the processing time. The flow time is the accumulation of processing time each job by each job. 4.3.3Sequencing Rules: Earliest Due Date (EDD)

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The formulas for calculation are below: #Average completion time = Total flow time ÷ No. of jobs #Average number of jobs = Total flow time ÷ Total job work time In the system #Average job lateness = Total late days ÷ No. of jobs #Utilization = Total job work time÷ Total flow time = in % 4.3.3Sequencing Rules: Earliest Due Date (EDD)

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Five engine blocks are waiting for processing. The processing times have been estimated. Expected completion times have been agreed. The table shows the processing time and due date of those 5 engines. Determine the schedule by using the EDD rule. Engine BlockProcessing Time (Days) Due Date (Days) Ranger810 Explorer612 Bronco1520 Econoline 150318 Thunderbird1222 4.3.3Sequencing Rules: Earliest Due Date (EDD) Example 1

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Engine (1) (2) (3)(2)-(3) Block Processing Completion Due Days Tardy Sequence Time Time Date(0 if negative) Ranger 810 Explorer 612 Econoline 150 318 Bronco 1520 Thunderbird 1222 80808080 3212 142 170 4422 Total 44 85 36 4.3.3Sequencing Rules: Earliest Due Date (EDD) Example 1 Average completion time85/5 = 17 days Average number of jobs in the system85/44 = 1.9318 jobs Average tardiness36/5 = 7.2 days Utilization44/85 = 51.76%

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Is an index number computed by dividing the time remaining until due date by the work time remaining. The critical ratio gives priority to jobs that must be done to keep shipping on schedule. The critical ratio is measure of urgency of any order compared to the other orders for the same facility. The ratio is based on when the completed order is required and how much time is required to complete. 4.3.4Sequencing Rules: Critical Ratio (CR)

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The step for using this rule are: 1.At the starting program, user input the numbers of job, the jobs name, the works day remaining and the due date of each job and as well the today's date. 2.The today's date and the number of job are just inputted once time. Then, the others are followed the value of the number of jobs inputted. After that, compute the critical ratio by using the formula. 3.The formula for Critical Ratio is: CR = time remaining / works day remaining 4.After calculating the CR for each job, give the priority order by using the value of the calculated critical ratio. The priority order is performed from smaller to larger. 4.3.4Sequencing Rules: Critical Ratio (CR)

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There are 3 characteristics can be seen from the critical ratio: A job with low critical ratio(less than 1.0) ---- falling behind schedule. If CR is exactly 1.0 ---- the job is on schedule. If CR is greater than 1.0 ---- the job is ahead of schedule and has some slack. 4.3.4Sequencing Rules: Critical Ratio (CR)

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The critical ratio help in most production scheduling system as below: Determine the status of specific job. Establish relative priority among jobs on a common basis. Relate both stock and make-to-order jobs on a common basis. Adjust priorities (and revise schedules) automatically for changes in both demand and job progress. Dynamically track job progress and location. 4.3.4Sequencing Rules: Critical Ratio (CR)

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A machine center in a job shop for a local fabrication company has five unprocessed jobs remaining at a particular point in time. The jobs are labeled 1, 2, 3, 4, and 5 in the order that they entered the shop. The respective processing times and due dates are given in the table below. Sequence the 5 jobs by CR rules. Job numberProcessing TimeDue Date 1234512345 11 29 31 1 2 61 45 31 33 32 4.3.4Sequencing Rules: Critical Ratio (CR) Example 1

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Current time: t=0 Job numberProcessing TimeDue DateCritical Ratio 1234512345 11 29 31 1 2 61 45 31 33 32 61/11(5.545) 45/29(1.552) 31/31(1.000) 33/1 (33.00) 32/2 (16.00) Current time: t=31 Job numberProcessing TimeDue Date-Current TimeCritical Ratio 12451245 11 29 1 2 30 14 2 1 30/11(2.727) 14/29(0.483) 2/1 (2.000) 1/2 (0.500) Current time should be reset after scheduling one job 4.3.4Sequencing Rules: Critical Ratio (CR) Example 1

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Current time=60 Job numberProcessing TimeDue Date- Current Time Critical Ratio 145145 11 1 2 1 -27 -28 1/11(0.0909) -27/1<0 -28/2<0 Both Jobs 4 and 5 are later, however Job 4 has shorter processing time and thus is scheduled first; Finally, job 1 is scheduled last. 4.3.4Sequencing Rules: Critical Ratio (CR) Example 1

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Average completion time289/5 = 57.8 days Average number of jobs in the system289/74 = 3.905 jobs Average tardiness87/5 = 17.4 days Utilization74/289 = 25.61% Job numberProcessing Time Completion Time Due DateTardiness 3245132451 31 29 1 2 11 31 60 61 63 74 31 45 33 32 61 0 15 28 31 13 Totals7428987 4.3.4Sequencing Rules: Critical Ratio (CR) Example 1

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4 Rules Application - Example Job numberProcessing TimeDue Date ABCDEFABCDEF 2 8 4 10 5 12 7 16 4 17 15 18 Processing Time (including setup times) and due dates for six jobs waiting to be processed at a work center are given in the following table. Determine the sequence of jobs, the average flow time, average tardiness, and number of jobs at the work center, for each of these rules: FCFS SPT EDD CR

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4 Rules Application – Example (FCFS) Job Sequence Processing Time Flow TimeDue DateTardiness ABCDEFABCDEF 2 8 4 10 5 12 2 10 14 24 29 41 7 16 4 17 15 18 0 10 7 14 23 Totals4112054 Average completion time120/6 = 20 days Average number of jobs in the system120/41 = 2.93 jobs Average tardiness54/6 = 9 days Utilization41/120 = 34.17%

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4 Rules Application – Example (SPT) Job Sequence Processing Time Flow TimeDue DateTardiness ACEBDFACEBDF 2 4 5 8 10 12 2 6 11 19 29 41 7 4 15 16 17 18 0 2 0 3 12 23 Totals4110840 Average completion time108/6 = 18 days Average number of jobs in the system108/41 = 2.63 jobs Average tardiness40/6 = 6.67 days Utilization41/108 = 37.96%

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4 Rules Application – Example (EDD) Job Sequence Processing Time Flow TimeDue DateTardiness CAEBDFCAEBDF 4 2 5 8 10 12 4 6 11 19 29 41 4 7 15 16 17 18 0 3 12 23 Totals4111038 Average completion time110/6 = 18.33 days Average number of jobs in the system110/41 = 2.68 jobs Average tardiness38/6 = 6.33 days Utilization41/110 = 37.27%

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4 Rules Application – Example (CR) Job SequenceProcessing TimeDue DateCritical Ratio Calculation ABCDEFABCDEF 2 8 4 10 5 12 7 16 4 17 15 18 (7-0) / 2 = 3.5 (16-0) / 8 = 2.0 (4-0) / 4 = 1.0 (Lowest) (17-0) / 10 = 1.7 (15-0) / 5 = 3.0 (18-0) / 12 = 1.5 At t=0, Job C is the first job to complete base on the lowest critical ratio.

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4 Rules Application – Example (CR) Job SequenceProcessing TimeDue DateCritical Ratio Calculation ABCDEFABCDEF 2 8 - 10 5 12 7 16 - 17 15 18 (7-4) / 2 = 1.5 (16-4) / 8 = 1.5 - (17-4) / 10 = 1.3 (15-4) / 5 = 2.2 (18-4) / 12 = 1.17 (Lowest) At t=4, day 4 [C completed], Job F is the second job to complete base on the lowest critical ratio.

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4 Rules Application – Example (CR) Job SequenceProcessing TimeDue DateCritical Ratio Calculation ABCDEFABCDEF 2 8 - 10 5 - 7 16 - 17 15 - (7-16) / 2 = -4.5 (Lowest) (16-16) / 8 = 0 - (17-16) / 10 = 0.1 (15-16) / 5 = -0.2 - At t=16, day 16 [C and F completed], Job A is the third job to complete base on the lowest critical ratio.

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4 Rules Application – Example (CR) Job SequenceProcessing TimeDue DateCritical Ratio Calculation ABCDEFABCDEF - 8 - 10 5 - 16 - 17 15 - (16-18) / 8 = -0.25 - (17-18) / 10 = -0.10 (15-18) / 5 = -0.60 (Lowest) - At t=18, day 18 [C, F and A completed], Job E is the fourth job to complete base on the lowest critical ratio.

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4 Rules Application – Example (CR) Job SequenceProcessing TimeDue DateCritical Ratio Calculation ABCDEFABCDEF - 8 - 10 - 16 - 17 - (16-23) / 8 = -0.875 (Lowest) - (17-23) / 10 = -0.60 - At t=23, day 23 [C, F, A and E completed], Job B is the fifth job to complete base on the lowest critical ratio and follow by Job D in last.

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4 Rules Application – Example (CR) Job Sequence Processing Time Flow TimeDue DateTardiness CFAEBDCFAEBD 4 12 2 5 8 10 4 16 18 23 31 41 4 18 7 15 16 17 0 11 8 15 24 Totals4113358 Average completion time133/6 = 22.17 days Average number of jobs in the system133/41 = 3.24 jobs Average tardiness58/6 = 9.67 days Utilization41/133 = 30.83%

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RulesAverage Flow Time (days) Average Tardiness (days) Average Number of Jobs at the Work Center Utilization (%) FCFS SPT EDD CR 20.00 18.00 18.33 22.17 9.00 6.67 6.33 9.67 2.93 2.63 2.68 3.24 34.17 37.96 37.27 30.83 4 Rules Application – Example (CR)

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